$\dfrac{d}{dx}\left(\dfrac{1}{x^4}+\dfrac{1}{x^2}-x^3\right)=$
Answer: The strategy We can first rewrite each rational term in the expression as a negative power of $x$. Then, the derivatives of these terms can be found using the power rule : $\dfrac{d}{dx}(x^n)=n\cdot x^{n-1}$ (Remember that this applies even when $n$ is negative.) Rewriting rational terms as negative powers $\begin{aligned} &\phantom{=}\dfrac{1}{x^4}+\dfrac{1}{x^2}-x^3 \\\\ &=x^{-4}+x^{-2}-x^3 \end{aligned}$ Differentiating using the power rule $\begin{aligned} &\phantom{=}\dfrac{d}{dx}(x^{-4}+x^{-2}-x^3) \\\\ &=\dfrac{d}{dx}(x^{-4})+\dfrac{d}{dx}(x^{-2})-\dfrac{d}{dx}(x^3) \\\\ &=-4x^{-5}+(-2)x^{-3}-3x^2 \\\\ &=-4x^{-5}-2x^{-3}-3x^2 \\\\ &=-\dfrac{4}{x^5}-\dfrac{2}{x^3}-3x^2 \end{aligned}$ In conclusion, $\dfrac{d}{dx}\left(\dfrac{1}{x^4}+\dfrac{1}{x^2}-x^3\right)=-\dfrac{4}{x^5}-\dfrac{2}{x^3}-3x^2$.